Question

giải hệ

\(\left\{{}\begin{matrix}x+3y-xy=3\\x^2+y^2+xy=3\end{matrix}\right.\)

Answer 1

Ta có : \(\left\{{}\begin{matrix}x+3y-xy=3\\x^2+y^2+xy=3\end{matrix}\right.\)

Xét PT đầu : \(x+3y-xy=3\)

Chuyển vế được : \(\left(x-3\right)\left(1-y\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

TH 1 : x = 3 \(;x^2+y^2+xy=3\)

\(\Rightarrow9+y^2+3y=3\) \(\Leftrightarrow y^2+3y+6=0\) ( ***** ) vì :

\(y^2+3y+6=y^2+3y+\frac{9}{4}+\frac{15}{4}=\left(y+\frac{3}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\forall y\)

TH 2 : \(y=1;x^2+y^2+xy=3\)

\(\Rightarrow x^2+x+1=3\) \(\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\) ( t/m )

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